Asked by Anna
Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s2.
Answers
Answered by
Damon
The pressure is determined by the height of water in the tank.
P = rho g h
rho is about 1000 kg/m^3
g is 9.81
P is 50 *10^3 Pascals
solve for h in meters
then the radius of the tank is 4.3 meters and it is filled to depth h
volume of water = (1/6) pi h (3 rh^2 +h^2)
where rh is the radius at the water surface with depth h so rh = sqrt (2Rh-h^2)
then convert those m^3 to liters
P = rho g h
rho is about 1000 kg/m^3
g is 9.81
P is 50 *10^3 Pascals
solve for h in meters
then the radius of the tank is 4.3 meters and it is filled to depth h
volume of water = (1/6) pi h (3 rh^2 +h^2)
where rh is the radius at the water surface with depth h so rh = sqrt (2Rh-h^2)
then convert those m^3 to liters
Answered by
Enty
Stop trying to cheat at neopets, it ruins it for the people who actually do the math.
Answered by
carla
296,1
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